Given a locally free sheaf $M$ on $\mathbb{P}^2$ with $h^0(M)=1$. Is it true that we have $h^2(M)=0$ in this case?

I got this idea from Friedman's book "Algebraic Surfaces and holomorphic vector bundles". In Chapter 4, p.109, Ex. 4 he wrote: $h^0(Hom(V,V))=1$, by Serre duality $h^2(Hom(V,V))=h^0(Hom(V,V)\otimes K)=0$ since $K=O(-3)\subset O$.

But I don't see why $h^0(Hom(V,V)\otimes K)=0$ follows from $K=O(-3)\subset O$. I mean it could stil have dimension one or is it because $O(-3)$ has no global sections? Would $h^0(Hom(V,V)\otimes O(-i))=0$ still be true for $i=1,2$? Can one generalize to arbitrary locally free sheaves or is this only correct in this special case, i.e. $V$ stable?

${\bf Edit:}$ Given a simple sheaf $V$ on $\mathbb{P}^2$, Bjorn's answer shows $H^0(Hom(V,V)\otimes O(-i))=0$ for $i>0$ which can be written as $Hom(V,V(-i))=0$ for $i>0$. Can this be generalized?

For example given a sheaf of rings or algebras $R$ and a simple left $R$-module M, do we always have $Hom_R(M,M(-i))=0$ for $i>0$? Or do I need to be more careful in this case?

I remembered this question reading arxiv.org/abs/0810.0067, page 8, where such an equality shows up, without further explanation, so i thought the argumentation should carry over to this more general case.